【LeetCode】LeetCode——第19题：Remove Nth Node From End of List

19. Remove Nth Node From End of List

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Question
Editorial Solution

Total Accepted: 106851 Total
Submissions: 362053 Difficulty: Easy

Given a linked list, remove the nth node from the end of list and return its head.
For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

题目的大概意思是：给定一个单向链表，删除其倒数第n个结点。

这道题难度等级为：简单

思路：遍历链表的长度，然后再次遍历找到倒数第n个结点，将其删除。

注意：输入的链表是没有头结点的。

代码如下：

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* p = head;
ListNode* q = head;
int len = 0, cnt = 0;
while (p != NULL){ //表长
p = p->next;
++len;
}
if (len == n){return head->next;}
while(cnt < len - n - 1){
q = q->next;
++cnt;
}
(n == 1) ? (q->next = NULL) : (q->next = q->next->next);
return head;
}
};提交代码，AC时间为Runtime:4ms。